Question

match the appropriate value for the denominators of ( x^2 ), and ( y^2 ) when the following equation for an ellipse is re-written in standard form. ( 4x^2 + y^2 - 64 = 0 )
denominator for ( y^2 ) choose...
denominator for ( x^2 ) choose...
64
12
4
8
16
find the inverse of th ( \begin{bmatrix} 6 & -5 \\ 3 & -2 end{bmatrix} ) rix

Explanation:

Step1: Rewrite the ellipse equation

Start with the given equation \( 4x^2 + y^2 - 64 = 0 \). Add 64 to both sides:
\( 4x^2 + y^2 = 64 \)

Step2: Divide by 64 to get standard form

The standard form of an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Divide each term by 64:
\( \frac{4x^2}{64} + \frac{y^2}{64} = 1 \)

Step3: Simplify the \( x^2 \) term

Simplify \( \frac{4x^2}{64} \) by dividing numerator and denominator by 4:
\( \frac{x^2}{16} + \frac{y^2}{64} = 1 \)

From this, the denominator for \( y^2 \) is 64 and for \( x^2 \) is 16.

Answer:

Denominator for \( y^2 \): 64
Denominator for \( x^2 \): 16